text:
The stall velocity condition: Exactly what is stall velocity?
Question: if my stall velocity is 20m/s, then...
If the airplane is travelling at a velocity greater than or equal to the stall velocity, can one assume the plane will not stall?
Can one assume that for a take off distance of 750m, as long as the velocity is greater than 20, the plane won’t stall (assuming constant angle of attack when take off)
Is this correct, or am I wrong somewhere? I just want to know generally please, in relation to lift and stall.

Does this answer your question? How does stall depend on angle of attack but not speed?

Hi, I didn’t really want to relate to Angle of Attack, just wanted to know from velocity and take off length point of view please

Is the first answer to this question useful?  https://aviation.stackexchange.com/questions/27706/does-the-expression-stall-speed-have-a-definition

Related question.  If I'm going 20 m/s with no wing at all, will I always fly?  It's a good idea to exceed stall speed first, then apply elevator to create sufficient AOA to fly.  See the relationship between velocity and AoA in the Lift Equation.

Assuming constant AoA upon takeoff .... but if the AoA at the speed you are going is too great, stall. AoA cannot be removed.

I disagree with the question being closed. The OP is under the (very common) misconception that stall follows directly from low speed, but that's hardly a reason to close the question. Correcting a misconception is the very reason to ask questions about it.

Your proposal cannot be held to be "true" as a general statement.  For example, imagine that the plane has tricycle gear, and we extend the nose wheel by ten extra feet so that the plane sits at a nose high attitude of 45 degrees or more on the ground.  The wing will be at a greater-than-stall angle-of-attack no matter how fast it accelerates, so it will never be able to take off.  The same would happen if the landing gear were normal, but at the instant of rotation the pilot violently yanked the nose up to beyond the stall angle-of-attack and then kept the a-o-a constant at that angle.

I'm with @HiddenWindshield  The question includes a widely held mistaken assumption about stalling and the current answer is useful in correcting that.

Your confusion isn't clear at all.  What is it about the definition of the term "stall speed" that you don't understand?

As the angle of attack of an airfoil increases, the lift it generates increases -- at least, up to a point. Once it reaches that point, called the critical angle of attack, the lift begins decreasing again. This reduction in lift is called a stall.
However, airspeed also affects the amount of lift generated. The slower the airplane goes, the less lift is produced. Thus, if the pilot wants to maintain altitude while going slower, they need to increase the AoA to compensate for the reduced speed.
If the speed continues to deteriorate while the pilot is trying to maintain level flight, then the plane will eventually reach its critical AoA. The speed where that happens is called the "stall speed". But the stall only happens at that particular speed because the pilot is trying to stay level.  If the pilot is willing to go into a dive, it's possible to fly below the defined "stall speed" without stalling*. Conversely, if the pilot is pulling up, this temporarily increases the AoA, which may make the plane stall even if it's above its "stall speed".
So, to answer your question, no, your thinking is not correct. It is possible to stall the plane, even if it's traveling faster than 20 m/s. The length of the runway is completely irrelevant.
*At least temporarily. To maintain that speed without stalling, you'd need to keep the wings unloaded, which would imply an ever-increasing rate of descent, resulting in the plane accelerating due to gravity. But you could, for a few moments, fly below "stall speed" without actually stalling.

I’d say instead: “If the pilot is willing to lose altitude, it’s possible to fly below the stall speed.” Some planes just “mush” at AoAs above critical.  But I would not say without stalling; they are “in the stall…”

@Jim A plane "mushes" when part of the wing has stalled, but part of it is still flying normally. And, yes, it's possible to fly indefinitely in that condition. But it is also possible, by pushing forward (increasing the rate of descent), to fly below stall speed without any part of the wing stalled (i.e. without "mushing"). Granted, you can't stay like that for very long since you'll start picking up speed from the dive, but it is physically possible.

Yes, that’s what’s happening during stall recovery. But it’s more of a brief transient condition.  I just thought your sentence was misleading because I took it to imply that you could reach a steady state of “flight” in that diving condition.   It’s also what’s happening on the runway as you are accelerating- you just have the ground to hold you up.

@Jim Good point, I added a footnote (since I couldn't figure out a clean way to put it in the actual sentence).


text:
I calculated that Flyer I and II (1903-04) of the Wright brothers were underpowered. Are there better (drag, lift) coefficients than those of Culick?
Based on what the Wright brothers said in a few letters and two article (all relevant excerpts have been attached to this post), and also using the experimental data in the study "AERODYNAMICS, STABILITY AND CONTROL OF THE 1903 WRIGHT FLYER - Fig. 12 Lift and Drag of the 1903 Flyer" by Fred Culick (1984), I reached the conclusions that both Flyer I and II were grossly underpowered and could not have reached the airspeeds claimed by the Wright brothers.
Are there better works than Culick's with more precise drag and lift coefficients?

These are my calculations:

Fragments, from letters and articles of the Wrights, relevant for the question:
1904-03-14, W. Wright, “Letter to O. Chanute”, Dayton, March 14, 1904
“We are hard at work getting ready for Spring. The new machines will be of the same size as the old one but will weigh a little more, 800 lbs. probably. By gearing the engine to run a little faster we will not only carry the additional weight but will have enough surplus to increase the speed to about 40 miles an hour.”
1904-05-20, W. Wright, “Letter to O. Chanute”, Dayton, May 20, 1904.
“Our indoor tests of the machinery show excellent results. With the same screws we used last year we get an increase in speed of 50 turns per minute, indicating an increase in power of more than one half. This is partly due to gearing the engine to run at higher speed per turn of screw, and partly to increase in efficiency of the engine itself.”
1904-06-14, W. Wright, “Letter to O. Chanute”, Dayton, June 14, 1904, 1 page.
“This machine is entirely new, including engine and machinery. We are using the old screws.”
1904-12-21, Orville Wright, “Letter to Carl Dienstbach”, December 21, 1904, 2 pages.
“The two longest flights of the season were made on the 9th of November and the 1st of December. In each of these flights we made almost four complete circles and covered a distance of a little over four and one half kilometers, at a speed of about 35 miles an hour. In the flight of November 9th a weight of 50 lbs. (iron bars) were carried in addition to the weight of the operator; in the flight of December 1st, 70 lbs.”
1907, Wilbur and Orville Wright, “The Relations of Weight, Speed, and Power of Flyers”, Navigating the Air - A Scientific Statement of the Progress of Aëronautical Science up to the Present Time - By the Aero Club of America, London, Heinemann, 1907, Printed in New York, U.S.A., pp. 6-12.
“THE flyer of 1903 carried a four-cylinder gasolene motor of four-inch bore and four-inch stroke. Complete with magneto, radiators, tanks, water, fuel, etc., the motor weighed a little over 200 lbs.; and at 1200 revolutions per minute, developed 16 horse-power for the first 15 seconds after starting. After a minute or two the power did not exceed 13 to 14 horse-power. At 1020 revolutions per minute — the speed of the motor in the flights at Kitty Hawk on the 17th of December, 1903, — it developed about 12 horse-power.
The flyer of 1904 was equipped with a motor similar to the first, but of 1/8 inch larger bore. This engine at 1500 revolutions per minute developed 24 horse-power for the first 15 seconds, but only 16 to 17 horse-power after a few minutes’ run. Complete with water, fuel, and other accessories, it weighed 240 lbs.
The same engine, with a few modifications in the oiling device and the carbureter, was used in all the flights of 1905. A test of its power made soon after the flights of October, 1905, revealed a gain of 3 horse-power over tests made just before mounting it on the flyer in 1904. This gain is attributed to the increased smoothness of the cylinders and pistons produced by wear. The small output of these engines was due to lack of experience in building gasolene motors. …
A comparison of the flyers of 1903, 1904, and 1905 shows some interesting facts. The flyer of 1903 weighed, complete with operator, 745 lbs. Its longest flight was of 59 seconds’ duration with a speed of 30 miles an hour and an expenditure of 12 horse-power. The flyer of 1904 weighed about 900 lbs., including a load of 70 lbs. in iron bars. A speed of more than 34 miles an hour was maintained for a distance of 3 miles with an expenditure of 17 horse-power. The flyer of 1905 weighed, including load, 925 lbs. With an expenditure of 19 to 20 horse-power it traveled over 24 miles at a speed of more than 38 miles an hour. The flights of 1904 and 1905 would have been slightly faster had they been made in a straight line, as were those of 1903.
In 1903, 62 lbs. per horse-power were carried at a speed of 30 miles an hour; in 1904, 53 lbs. at 34 miles an hour; and in 1905, 46 lbs. at 38 miles an hour. It will be noticed that the weight carried per horse-power is almost exactly in inverse ratio to the speed, as theory demands — the higher the speed, the smaller the weight carried per horse-power.
Since flyers can be built with approximately the same dynamic efficiency for all speeds up to 60 miles an hour, a flyer designed to carry a total weight of 745 lbs. at 20 miles an hour would require only 8 horse-power, or two-thirds of the power necessary for 30 miles an hour. At 60 miles 24 horse-power would be necessary — twice that required to carry the same weight at 30 miles an hour. At 120 miles an hour 60 to 75 horse-power would probably be necessary, and the weight carried per horse-power would be only 10 or 12 lbs. At such high speed the resistance of the operator's body and the engine is a formidable factor, consuming 64 times as much horse-power as at 30 miles an hour. At speeds below 60 miles an hour this resistance is almost negligible.”
1908-09, Orville and Wilbur Wright, “The Wright Brothers’ Aeroplane”, The Century Magazine, New York, September 1908, Vol. LXXVI, No. 5, pp. 641-650 (pp. 648-649).
“Our first propellers, built entirely from calculation, gave in useful work 66 per cent of the power expended. This was about one third more than had been secured by Maxim or Langley.”

Flyer I, Kitty Hawk, North Carolina, December 17, 1903.

The Wright flyers at Kitty Hawk may have been helped a bit by rising airflow from the sea breeze against the dunes. No doubt they were marginally powered, noting that an increase in AOA from 5 to 10 degrees could more than double the drag coefficient.  One can see how slightly higher airspeed (lower AOA for the same lift) was an advantage.  Little wonder they designed their planes to "belly flop" when they "stalled". See Gustave Whitehead, another early pioneer.

I think the Wrights made a big point of making sure the powered flights were on flat ground; I don't think they benefited from slope lift.

"The 1903 plane was considerably overpowered if we are to trust the results of the article I quoted." -- did mean to write "underpowered"?

@quietflyer, You quoted a text written by somebody else, in his answer. He concluded, based on the data he found, that Flyer I was overpowered.

The results of a trial made on November 21, 1903, do not match the 1999 wind tunnel measurements.
The discrepancy is enormous. The Wright brothers and the people at NASA tested very different planes and propellers otherwise the reported experimental data would have matched well.
I do not agree the 1999 experiments, on the full size model of the 1903 plane, gave more precise measurements than the ones obtained by Fred Culick and reported in 1984.
From Full-Scale 1903 Wright Flyer Wind Tunnel Test Results From the NASA Ames Research Center - 1999, page 18:

In the wind tunnel, the reproduced propeller fell short of T=D @ 28
mph test condition when operating at the maximum permitted speed of
340 RPM. These data are being analyzed

The same presentation, at page 10, contains a measurement for $RPM=340$ showing $C_D=0.035$ and $C_L=0.66$ (I extracted these precise values from the diagram using a special software tool).
The minimum drag at 28 mph should have been:
$$D=W\frac{C_D}{C_L}=745lbf\frac{0.035}{0.66}=39.50lbf,\space(V_{air}=28mph, RPM=340)$$
$$P_{340RPM, \space28mph}=\frac{V_{air}D}{\eta}=4.47hp, \space \eta=0.66$$
From Orville Wright's 1903 Notebook:

Saturday, Nov. 21  After many attempts to fasten sprockets we finally
succeeded by filling threads with tire cement. The engine ran very
irregularly, jerking the chains, and shaking the machine terribly. We
discovered the trouble lay in the gasoline feed, and after fixing
valve so that the vibration could not change it we had no further
trouble from that source. The first test of speed was 306 rev. screw
to minute (309 in 60½ sec). One cylinder made only a few explosions
during the test. On the next trial we got 333 rev. in 60 sec. After
dinner we arranged to measure the thrust by supporting center skids on
rollers and fastening one end of machine, while we attached a rope to
the other end, which ran over a pulley and carried a 50 lb. box of
sand. Besides lifting the sand we got an additional pull of 16 to 18
lbs on the scales, which made the total thrust of the screws 132 to
136 lbs, at a speed of 350 revolutions per min. Our confidence in the
success of the machine is now greater than ever before. Weather today
was warm and pleasant with 6 to 8 meter wind from North East.

Orville Wright says that in a wind of $6$ to $8 m/s$ (I will take $V_{air}$=$7 m/s = 15.69 mph$ on average), at $RPM=350$, the Wrights got the thrust:
$$T = 132 \space to \space 136 lbf, \space (V_{air}=15.69mph, RPM=350)$$
It is true , there is a difference of $10 RPM$ between the test described by Orville and that made in the wind tunnel but this can not explain the enormous difference between the thrust obtained by the Wright brothers $T_{min}=132lbf$, which would have further increased at $V_{air}=28mph$, and that of the group of experimenters, at the NASA facility, which was evidently $T<39.50lbf$.

Only $P=4.47hp$ to fly a $745lb$ plane at $28mph$ seems unrealistic.

1. I have found this article:
The Wilmington Messenger, Wilmington, North Carolina, May 26, 1904, col. 1, p. 6.
Elizabeth City Economist: A gentleman visiting this city whose home is in Kitty Hawk, is responsible for the assertion that the Wright brothers, of airship fame, will return to Kitty Hawk in the near future and resume work on their aerial monster. According to this gentleman the airship has never been removed from Kitty Hawk and nearly all the interviews published in the papers of Norfolk have been erroneous in this respect. This gentleman has assisted the Wrights in all their work and has a general supervision of their property during their absence. He says that they have not completed the ship and that they will return some time within the next month and resume their work. A story is current that they will complete the ship and make the trip from here to St. Louis sometime this fall.
The text was published (what a coincidence!) in the very day, May 26, 1904, when the Wright brothers flew Flyer II, near Dayton, for the first time. The article says that according to a man that worked for them and took care of their things left behind at Kitty Hawk, the two inventors hadn't finished Flyer I. In conclusion, this machine did not fly on December 17, 1903!
2. Coming back to the question, yes, there are better experimental data than Fred Culick's. They can be found in the article: "Full-Scale 1903 Wright Flyer Wind Tunnel Test Results From the NASA Ames Research Center", from which I extracted the following diagrams:

Flyer I 1903 - Wind Tunnel Data - Effects of Power
As can be seen in the diagrams above, for 300 RPM, if $C_D=0.06$ then $C_L=0.7$, and using the formulas given in the question, it results that the power needed by Flyer I, on December 17, 1903, was only $P_{300}=7.74 hp$, which is ridiculously low.
For 340 RPM, at $C_D=0.04$ ($C_L=0.7$), the necessary power would have been just $P_{340}=5.16 hp$.
The 1903 plane was considerably overpowered if we are to trust the results of the article I quoted.
(The weight of the plane was taken as $W=745lbf$, speed $V=30mph$, propellers' efficiency $0.66$)
As a note: All experimental tests with models of various sizes, representing Flyer I, have as their primary source of data the 1916 replica built by Orville Wright, not the 1903 original.

Robert the Reynolds numbers of the models were slightly lower than the "real" Flyers which were around 1 to 2 million at 20-40 mph (9-18 m/s with a 2 meter chord).  This is right in the range (from polars) where higher Reynolds numbers will give much more favorable Cl/CD ratios, especially at lower angles of attack.  The Wrights work from kite flying and wind tunnels really paid off, albeit just barely.

Yes, the prop airflow did help.

"In conclusion, this machine did not fly on December 17, 1903" --!!

Fred Culick measurements and those of AIAA group agree for the case when the propellers did not run: "Comparisons (props-off) with previous wind tunnel tests of smaller models showed fairly close agreement" (see: http://www.wrightflyer.org/wp-content/uploads/2012/10/Full-Scale-1903-Wright-Flyer-Wind-Tunnel-Test-Results.pdf page 16).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Application interview / career as a pilot: is qualifying from First Officer to Captain expected from an applicant?
I would like to know, if an airline expects an applicant, who wants to become a pilot, to qualify from pilot (FO) to captain (CPT)? Is qualifying kind of mandatory or is it primarily a personal decision?
I am asking because I do scientific investigation on pilot assessment. So if someone can share personal experience or even official articles and papers, I would be grateful.
EDIT: I have to add the background info, that the airline company is the provider of the apprenticeship (i.e. Lufthansa in germany).

I’ve never heard of a First Officer position being considered an apprenticeship. After all, First Officers are trained, qualified pilots expected to be able to act as Pilot in Command at any point in time. In the US, everything about a pilot’s career is based on seniority. Because of this, some very qualified FOs choose to remain senior FOs in order to get the best schedules. Jeffrey Skiles is a perfect example of this. A better example of apprenticeship is the cadet programs of many airlines in clouding Lufthansa.

I would say that an airline will avoid hiring an applicant that is not assessed as eventual "captain material".  It's not the maximum return on the airline's investment.
An applicant would never say in an interview that they only wanted to be a lifelong FO, except under special circumstances like an older hire who is likely to retire before meeting the experience requirements, or someone who has limiting medical conditions at hire (like the Type 1 diabetics who are now flying in some countries) that prevent them from qualifying for PIC.
An FO has to apply to upgrade to captain, so theoretically, someone can continue as an FO indefinitely (in a union operation at least), up to a point, but I think it would be career suicide to say so up front during the interviews.
It does happen though.  An acquaintance of mine some years back was still copilot at a major airline, flying heavies, with somewhere around 25 years in.  He finally upgraded to captain for the last years of his career.  One of his motivations was he'd gotten to the point of being at the top of the seniority ladder for copilots.
He was making way less than he would as captain, but he didn't care about the money so much, and was more interested in the bidding benefits of being top of the heap in his "junior" heap in bidding for pairings, insofar as he got the best routes and the best schedules and the most time off.  Upgrading to capt would mean he had to drop to smaller aircraft and a lower rung on the seniority ladder(back to working over Christmas, etc), and temporarily less money .  He finally upgraded to capt more or less in response to management and peer pressure in his later years (he said at the time).
There are a number of websites that post interview Gouges ("gouge" is a slang word for intel reports on an airline's recruiting/interview practices) like here. These can give a good sense of the kind characteristics they are looking for, and you will never find one that advises people to state that they don't want to progress to captain.

Some people also stay as FO until they have the seniority to hold a line as CA (reserve sucks), or to get the base they want (commuting sucks), even if it means giving up some pay. Nobody would fault those choices, but as you say, they’re temporary.

Back in the day when the lines ran a bit less lean staff wise, bidding reserve at a major meant only flying occasionally.  Another guy I knew was a senior capt at a major and bid reserve all the time, because he only did about one overseas trip a month and was home running other affairs the other 3+ weeks. These guys all had businesses on the side because of all the time off.

I know guys on reserve now who need to pick up OT just to stay current. But that’s due to COVID, not the norm.

@JohnK - My understanding was that Jeffrey Skiles had his own full time home renovation/construction contracting business. He was flying as FO with over 20,000 flight hours.

A lot of them also do it to have a ready alternative to move to if they medical out.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Normal tendency of helicopter with counter clockwise rotating main rotors (top view) with force trim off and leaving controls?
Is it normal for the helicopter with CCW rotating main rotors to pitch down and roll left in forward flight, when force trim is turned off and hands off the cyclic?
Moreover, when we put the helicopter in right bank, would it tend to return to straight & level / left bank after a few seconds when hands off the cyclic (force trim off)?
By forward flight I mean 80 to 100 kts

Your title states "counter rotating" while your question states "counterclockwise rotating main rotors". Do you mean a single, CCW rotating main rotor, or do you mean dual counter-rotating rotors (plural)?  Please [edit] your post to clarify.

I meant CCW rotating rotor system (2 bladed to be specific)

With a American made helicopter equipped with a CCW rotating rotor, the helicopter will, as it accelerates past 10-15 KIAS, begin to pitch nose up as well as roll to the right.  This is a direct result of Effective Translational Lift (ETL) as well as Transverse Flow Effect.  Airflow through the main rotor disk past these speeds begins to take a more horizontal flow through the rotor disk, increasing lift on the advancing side of the rotor disk resulting in this.  European made helicopters with a CW turning main rotor will have a tendency to still pitch up as ETL becomes effective but will roll left due to gyroscopic precession from a greater lift imbalance on the front of the rotor disk being applied to the right side of the disk.

Both of the previous answers are really good, but if you're after a practical answer to your question I thought I'd see what happens in an EC135 (CCW rotor). With the Autostabilisation and back-up SAS disengaged flying straight & level at 90kts the aircraft is quite unstable. As such, whatever slight attitude the aircraft is in prior to loosing the controls, it continues in that direction. This departure is very rapid but controllable, there is no bias in any particular direction. This also applies in turns, the attitude very quickly continues in the direction of the control input. Hope that helps.

The answer depends on what disturbs the helicopter (with no disturbance it has no tendency).  For example, let's say there's a gust of headwind, blowing into the nose of the aircraft, increasing the airspeed.  This adds more lift to the advancing side of the main rotor (pilot's right side for CCW), which has the effect of flapping the rotor up 90 degrees later over the nose and pitching the helicopter nose up.  This end effect (pitch up) is the same regardless of the direction of rotor rotation.
Depending on the rotor design, the largest increase in flapping may be less than 90 degrees, which would add a (very secondary) roll effect.  Here the direction of rotation matters.  For a CCW main rotor this would cause a "left wing down" roll moment.
For more about this 90 degree (or less for some rotor designs) flapping offset see this:  Basic Helicopter Flap Dynamics.
Another effect of the increased airspeed is that it focuses more of the wake in the rear of the rotor, decreasing lift there relative to the front of the rotor.  This causes the rotor to flap down 90 degrees later on the pilot's right side (CCW), which adds a "right wing down" roll moment.
Analogous comments apply to the right wing down bank turn.  Let's say the pilot hadn't trimmed the climb rate yet, he entered the turn but is now accelerating down due to the main rotor thrust being tilted further from vertical.  Since the rotor is tilted right side down, the increase in air flow up (due to the helicopter falling down faster) adds velocity and lift to the rear of the rotor (CCW) in the same way increased airspeed added lift to the right side of the rotor.  This lift will cause the right side of the rotor to flap up, pushing back towards a wings-level condition.
As you can see, in both cases, the primary response is stabilizing.  This is described better here: Helicopter Stability


text:
increasing lift in a plane by placing engine in such a way that the air coming out of it washes over the wings to create more lift
I was wondering if the engines of a plane can be placed in such a way so that the air coming out of the engines goes from over the wings so that the velocity of air over the wing is higher and more lift is generated. has this ever been done? if not than why?and what are the pros and cons of this design?
i mean something like this

Check out the Boeing YC-14.

@ Fred Larson thanks. Do you know why so fewer planes like this exist?

I don't have a good answer for that, but there are folks here who may. Most likely there are tradeoffs that don't work out that well.

Related question that my give some insights: https://aviation.stackexchange.com/q/50127/520

Pretty much all the big turboprops, with their enormous 12 ft propellers covering nearly a 3rd of the wing span, enjoy this effect.  When flying an approach in a Dash 8 or similar, power changes have a near instantaneous effect on sink rate.

@JohnK ya but the accelerate air both above and below the wings

All the better.  Lift is mostly from the air mass below and above the wing being directed downward by the wing's passage; the "suction" of the low pressure zone is just a part of the whole effect.  Adding energy to the entire flow field will have a bigger effect than just the top flow.

https://aviation.stackexchange.com/q/82163/53144 answered here

I would think one reason is that jet exhausts are hot.  Aluminum alloys - to say nothing of composites! - don't really like high temperatures.


text:
Engine out over a mountain road. Do you land upwind or up hill?
You're flying solo in a late-model, fixed-gear Cessna 172 with half-full fuel tanks.  You are at 1000 feet AGL, 100 knots indicated, directly above the middle of a 2000 foot long straight section of paved road.  There is a steady 15 knot wind blowing up the 10 degree, wide, flat and clear slope that the road runs straight up.  Your engine suddenly seizes, prop horizontal, with no possibility of restarting it.  Everything else about the aircraft is perfectly functional and the height of the terrain is not significantly above sea level for performance purposes.  The road section heading is 360 degrees, due north up the hill.  Choose any initial heading for the engine failure and remember it occurs directly over the mid-point of the road section (runway).  Please provide information on the specific flight segments in terms of speed, turn rate, descent rate, etc. and consider the effects of the wind blowing directly upslope (without any crosswind component with respect to the roadway).  Explain why your answer is (nearly) optimal.  How do you make the safe landing?

It seems essential to give some indication of the actual slope (gradient) in mathematical terms.

Just to be sure: 10 degrees is the slope of the road (and terrain of course)? And: what is my heading compared to road: parallel, perpendicular, arbitrary?

Just mentioning that most people would refer to a 10 degree slope road as "a cliff". It is ridiculously steep. You would need to drop your glide slope down to 5.5:1 to just maintain altitude, going downhill.

@PcMan  Heh, yes it is somewhat steep for a "runway".  I calculate the glide slope to maintain the same height AGL as closer to 5.7 to 1.  But at near stall, power off, and 30 degrees flaps with a headwind, your glideslope with respect to the ground is closer to 4:1.  And skiers would refer to a 10 degree slope as a "bunny hill".  ;-)

@OCPatch if you work the potential energy mgh vs kinetic energy 1/2mv$^2$ your uphill approach looks good indeed, and supports what they do at the Wasserkuppe.  Keep going on this, as it is worth many upvotes.

Admittedly, the 10 degree sloped roadway is an anomaly, and you would never see an actual runway built on this grade.  Don’t try to apply any of this logic to a shallower runway without some careful analysis.
TLDR:  I favor the uphill/downwind landing because it gives you more time for maneuvering and requires less pilot skill to manage the roll-out after landing.
Here's my take on the possible landing approaches.  I think my performance estimates are generally conservative.  I think the flight maneuvers are fairly pedestrian until landing and roll-out.  Your mileage may vary.  These are not optimal, but I think it is possible to land either uphill or downhill with varying degrees of risk and maneuver aggressiveness from nearly any initial heading.  Obviously, some headings significantly reduce the risk of one direction over the other.  In half of all cases you may have to make right turns instead of the depicted left turns.
Here are my assumptions:
Altitudes are all referenced to the roadway midpoint elevation
Aircraft weight is 2000 lbs.
40 KIAS -- stall in landing configuration
45 KIAS -- 1320 ft/min descent - 30 degrees flaps
48 KIAS -- stall in clean configuration
50 KIAS  -- 780 ft/min descent - 10 to 20 degrees flaps
55 KIAS  -- 725 ft/min descent
60 KIAS  -- 800 ft/min descent
65 KIAS  -- 870 ft/min descent - best glide angle through the air  approx. 7.5 to 1 glideslope
70 KIAS  -- 970 ft/min descent
75 KIAS  - 1115 ft/min descent
80 KIAS - 1285 ft/min descent
The wind is blowing from the South at 15 knots
Horizontal wind component up the slope:  ~1500 ft/min
Vertical wind component (upward) up the slope:  ~264 ft/min
Uphill landing

Using your cat-like reflexes and finely-honed piloting skills, track outbound on a heading of 235 for 16 seconds while trading your 100kts of airspeed for a little more altitude.  I calculate you can pop up about another 80 feet from the mid-field starting altitude during that time.
Fly for an indicated airspeed of 70 kts because you must now turn (~1.1G load factor) at 6 degrees per second to a heading of 180.  This takes about 9 seconds; you are 1600 feet south, 2280 ft west, and about 950 ft above mid-field.  26 seconds have elapsed.
Fly on a heading of 180 for nearly 40 seconds.  Think about making a radio call stating position and situation.  Maintain 70 KIAS for better penetration into the wind.  You are in your downwind leg and it has been over 1 minute since your engine failed.  You are 5230 ft south, 2280 ft west, and 495 ft above mid-field.  About 65 seconds have elapsed.
Now turn, maintaining 70 KIAS, at 6 degrees per second to the "runway" heading of 360.  Take a few extra seconds to roll out and line up with the roadway.  You are about 3935 ft south (thanks to that upslope wind), aligned with the centerline of the road, and 108 ft above midfield (about 750 ft AGL because of the slope).  About 97 seconds have elapsed.
Slow to 65 KIAS and fly this heading for about 27 seconds.  Wait until the end to drop maybe 10 degrees flaps to preserve speed for the zoom-climb, flare landing.  Nose-high attitude will be your friend on the upslope landing so don't use too much flap deflection.  About 114 seconds have elapsed.
You cross over the landable portion of the road at 65 KIAS with over 100 ft AGL.  It has been about 118 seconds since your engine failed.  Fly the aircraft through an energy-preserving flare (actually climbing at the last) until a few feet above the rapidly rising road.  You will touch down just past the first third of the landable roadway a little over 2 minutes since your engine failed.  Your speed will bleed off, the airplane will stop flying, and because of the upslope, you will roll to a stop, with or without brakes in 10-20 seconds.  Apply brakes so you don't roll backward down the hill!
Downhill landing

Track outbound from the midfield point on a heading of 90 degrees for 7 seconds while trading airspeed for altitude.  The maneuver takes about 15 seconds to complete, but half-way through, begin a 6-degree-per-second turn, stabilizing at 70 KIAS and rolling out on a heading of 360.
Fly the 360 heading for about 4 seconds.  You are now about 2340 feet north, 2255 ft east, and about 925 ft above the midfield point.  26 seconds have elapsed.
Maintaining 70 KIAS, begin a 30-second, 6 degrees-per-second turn to heading 180.  You are now about 2940 ft north (upslope wind again), aligned with the roadway, and 550 ft above the middle of the roadway.  Your height AGL is only about 65 ft (!), but that’s okay.  You are on final approach and you can fly into and out of ground effect as you like to fly to the threshold of the landable portion of the roadway which is on the steeply descending slope.   It has been about 56 seconds since your engine failed.
Just under 60 seconds have elapsed.  At this point you need to increase your rate of descent to more than the 10 degrees (17% gradient) downslope of the hill and roadway.  You might choose to drop 30 degrees flaps and decrease airspeed to 45 KIAS and plunk down on the roadway at about 30 knots groundspeed, but you have to bleed off 35 knots of airspeed without ballooning too much and end up at only about 5 knots above stall.  Your angle of attack would be so high that you would not be able to see the roadway at all - not even the far end because of the down slope. And you would strike the tail on the ground before the main wheels touched.
Instead, you will need to increase your airspeed to about 80 KIAS and use about 30 degrees of flaps to keep from flying up and away from the ground.  This feels like a fairly steep approach but you are only descending at about 250 ft/min with respect to the ground which is falling away steeply.  You are about 2000 feet from the landable portion of the runway.
In about 20 seconds you are down and on the roadway about 100 yards down from the start of the landable portion.  About 1 minute and 17 seconds have elapsed since your engine failed.
Here's where things start to get a little sketchy.
You are rolling down the roadway with 80 knots indicated (65 kts ground speed) – give yourself 6-8 seconds to get the flaps fully retracted while you try to keep the aircraft “stuck” on the ground.  On a 10 degree downslope your brakes are (cosine 10 degrees) still about 98% as effective, if you weren’t still generating a lot of lift.  Conservatively, you might get 50% of your weight onto the wheels initially.  And don’t forget your mass is effectively pulling you down the slope (sine 10 degrees * 2000 lbs) with about 345 pounds of force.
For reference, this source: (https://aviation.stackexchange.com/questions/67979/is-there-an-actual-rated-thrust-for-a-cessna-172-engine-at-takeoff-power#:~:text=As%20John%20K%20said%2C%20it,be%20aware%20of%20before%20takeof) puts the thrust of the 160hp, fixed pitch 172N at between 500-600 lbs force when at takeoff power.
You will trade lift at higher speeds of the roll-out, which will make your brakes less effective, for the drag at the higher speeds which will counter-act the downhill force.
It’s probably a 30+ second, white-knuckle ride down the roadway managing your energy, trying to keep the plane “stuck” on the ground and to not lock up the lightly loaded mainwheel brakes until you get it stopped just before the end of the useable portion.
I conclude that the uphill/downwind landing is the better choice for the following reasons:

Time.  Compared to the downhill/upwind landing you have about 30 seconds more seconds to collect your thoughts, take calming breaths, and possibly make a radio call reporting your position and situation on one of the straight legs of your “pattern”.  You buy the extra time by landing on the end of the roadway that is almost 350 feet lower than the top end.
Distance.  Right up until touchdown, the proximity of the ground is not a factor at any part of the uphill/downwind approach.  Contrast this with virtually skimming the ground for 15 seconds at the last of the downhill final.
Energy.  Granted, change in energy on the uphill approach happens much more quickly at the landing, but because you are flying up the slope there is some leeway on the timing and aggressiveness of the landing flare.  And once you are down, you will roll to a stop even if you don’t use the brakes.  On the downhill landing both your kinetic energy and your potential energy are working against you getting your aircraft stopped.  The relatively feeble upslope wind does little to help you here, and in fact, hurts you once down by producing more lift and reducing the weight on wheels for braking.  There will likely be an excruciating 30 seconds of managing flaps, brakes, and elevator so you can brake to a stop before the end of the useable roadway.

'''''''''''''''''''''''''''''''''''''''''''
UPDATE:  Long edit coming that verifies the downhill/upwind landing is a viable option.
TLDR:  @Robert DiGiovanni has a really good point about using forward slips to control speed/energy in steep descents.
It's been more than a month but I finally got a chance to go flying and experiment with forward slips.  We had three people in the "steam-gauge" panel Cessna 172S, including a very experienced instructor pilot from one of the local FBO's. 3/4 tanks on both sides had us at about 2450 lbs on the ramp.  We didn't get as much data as I had hoped due to an abundance of caution and other pressing purposes for the flight, but here's what we found:
60 KIAS:
30 degrees flaps, steady-state, idle power glide - about 650-700 ft/min descent
20 degrees flaps, full rudder forward slip, ailerons to maintain heading - about 1000 ft/min descent
30 degrees flaps, full rudder forward slip, ailerons to maintain heading - about 1300 ft/min descent
The rest of these cases are 30 degrees flaps, full rudder forward slip, ailerons to maintain heading
65 KIAS:  ~1600 ft/min descent
70 KIAS:  ~1800 ft/min descent
75 KIAS:  2000+ ft/min descent
80 KIAS:  Not brave (or foolish?) enough to fly this close to the top of the white arc in a full slip where airspeed readings are inaccurate and delayed!
We often could not establish (or accurately sustain) exact airspeeds, and also, as we all know on some level, the response of the airspeed indicator and vertical speed indicator are both inaccurate and significantly lag (~7 seconds for our VSI) the actual aircraft state, especially in a full rudder slip.
I tried to use (potential and kinetic) energy analysis methods to compensate for some of the lag but the data set was small enough and noisy enough that I judged it better to estimate the time lag and take appropriately delayed readings from the vertical speed indicator.  This said, even though the VSI read 0 on the ground, I couldn't trust it as 100% accurate - so take all this data with a grain of salt.
Other things learned:
This airplane doesn't really like to be in a full (30 deg) flaps forward slip.  At one point the airspeed indicator read 60 KIAS and the stall horn was intermittently beeping (turbulent air over the port?).  In another prolonged slip, with well over half in both tanks, the low fuel indicator blinked a couple of times for the higher trailing wing.
I was also surprised to see the vertical speed indicator read only about 500 ft/min descent at 68-70 KIAS throughout a standard rate turn.  I verified via my video of the panel that we descended about 800 ft in one minute in the mostly steady-state turn.  It should theoretically be more like 900 ft/min in still air.  The air seemed smooth, but we may have had weak, broad lift over the large decommissioned airbase below us - and maybe the vertical speed indicator was no longer well-calibrated.
So to sum up, there were lots of potential sources of errors, but the data held no big surprises.  It verified @Robert DiGiovanni was right that you can get controlled, high rates of descent by using forward slips.  The downhill/upwind landing becomes much more manageable.  The 65 KIAS, 30 degrees flaps case he suggests gets you a descent rate into the headwind that gives you a 15.3 degree descent angle.  If you can wrap your head around this, use this like a (15.3 - 10) 5.3 degree glideslope or less to get to the threshold.  Use less than full rudder slip for a shallower approach angle and the eventual 'flare'.  If you can keep the airplane stuck on the ground it was the equivalent of a ~50 knot touchdown onto the road surface (but going downhill though), not the 80 kts that my approach case got you.  I would still pick the other direction, but the downhill/upwind option would definitely work, and in the hands of a fairly skilled pilot, leave you an airplane without a scratch on it.
As for that seized engine that got you here in the first place?  That's a problem for when you get the airplane back off the mountain road and into a mechanic's shop.

@OC Patch this is very well written. 1. Talk to Bush pilots. 2. Try some forward slips in the 172.  You'll be amazed at how steep you get.  "Landing AoA so high you can't see the runway", no, your flaps are at 30 and you're landing relative to the runway.  Your gravity thrust vector analysis is excellent, so we wind up trying to land on a flat runway at around 1500 rpm.  Hmm, gotta add some drag.  I would try a full forward slip at 65 knots (don't speed up to 80, just kill lift by reducing AOA).  Remember, your groundspeed is 35-45, which is survivable even if you go all the way to the bottom.

@Robert DiGiovanni You make a really good point about the forward slip.  The additional drag from the slip is continuously variable from zero all the way up to the point where you run out of rudder or aileron authority (maybe elevator?).  If anyone can give me tips or data on how to model that forward slip drag effect I will let the group know what I find.  Any Cessna/Textron engineers out there?  Sounds like a good excuse to go flying otherwise. ;-)

absolutely!  All this can be tried at altitude.  Now let's add angle of descent to your excellent rate of descent charts for head wind, no wind, and tail wind!

Plugging in arctan for 8knot descent rate/(55-15) you make 10 degree landing, but may have an issue with bouncing.  Very little doubt I would most likely go to the bottom and maybe stop, but it takes out the fatal 80 knot overshoot.  Time over a 2000 foot runway would be do or die in 20 seconds or less.  I think I would take the 30 knot dinger unless I really practiced the uphill.

Here's an uphill grass field landing in the Pyrenees Mountains of France by a low-wing light plane with small diameter tires and wheel pants (!).  ( https://www.youtube.com/watch?v=Tm7l11PGldI )

Unless the wind is too strong you should try to land uphill.
Tailwind will increase  the required length of the runway.
When you land uphill you have a significantly larger angle between your glide path and the hill slope compared to landing on a horizontal runway. That means the required length of the runway decreases.
Both effects work against each other so that (depending on the strength of the wind) the resulting required length of the runway is approximately the same as landing on a horizontal runway with no wind.
When you land downhill you have a significantly smaller angle between your glide path and the hill slope compared to landing on a horizontal runway. That means the required length of the runway increases significantly.
When additionally landing against wind downhill
the oncoming wind reduces the ground speed and thus increases the angle between your glide path and the hill slope.
But the wind coming uphill also generates additional lift (known as ridge lift) that flattens your glide path even more and thus the angle between the hill slope and  your  glide path is decreased even more so that it is questionable that you even touchdown before you reach any obstacles at the end of the hill.
For weak and medium wind speeds the combined effect of down slope and ridge lift  is stronger that the tailwind effect at uphill landings (depending on the slope angle and wind speed).

Disclaimer This only applies if you cannot touchdown at the hill top and your touchdown point would be on the downhill slope.

Tailwind landings uphill are common on airfields with a sloped runway. One famous example is the Wasserkuppe air field in Germany:
This Youtube video shows a pattern flight at the Wasserkuppe airfield starting downhill against the wind and landing uphill with tailwind. When the pilot starts the engine on the right you see a glider landing downhill against the wind.

Please look at the groundspeed and amount of runway used up in the flare.  But I hear they do know what they're doing at the Wasserkuppe.

@RobertDiGiovanni Yes, we know. ;o) Even worse pilots from outside are only allowed to land uphill.

This ones airspeed  was 80!  Interesting is that one hits the threshold way fast, but climbs to slow down.  Gravity never fails.  I'd like to learn it.

Landing downhill/into the wind may result in more wear on the brakes if done often, but would be much safer as an emergency procedure, especially in unfamiliar terrain.
A 10 degree downslope on a road would be a 17% down grade, a very rare road indeed.  The glide slope of a 172 is around 1:8 which works out to a 7 degree downslope (using sin and arcsin trigonometry), which is a 12.5 % downgrade, what to do?
With an emergency landing, it is best to land into the wind.  A crash at 30 knots groundspeed is far more survivable then at 70 knots.  Because of weight considerations, the 172 is far more fragile in a collision than a modern car.  Walking away from a 70 knot crash is highly unlikely.
Keep in mind, in a steady state glide, the component of gravity pulling you forward, sine(glide angle) x mass, is equal to your drag and should not dissuade one from landing at a steep angle.
Gravity will only increase the roll-out distance as the plane slows down after touchdown, but this is offset by much lower ground speed when landing.
For the downwind/uphill landing drag and gravity now work together to rapidly slow the aircraft, but care must be taken as the groundspeed, therefor time over runway, is much less.  In this particular case, calculating 1/2mv$^2$ vs mgh at 65 knots (33 m/s) will work well:

Kinetic Energy from Speed = 900 kg/2 × (33m/s)$^2$ IAS = 490,000 kg (m/s)$^2$

Potential Energy from Climb
= 900 kg × 10m/s$^2$ × 100m = 900,000 kg (m/s)$^2$

even without factoring in the drag contribution:

Drag over Runway = 900kg/2 × 33(m/s)$^2$ approach speed - 900kg/2 × 23(m/s)$^2$ landing speed = 252,000 kg(m/s)$^2$

or ground effect.  As with all landings, too high an approach speed will carry the risk of overshooting the runway, with 15 knots tailwind added.
1000 feet AGL is plenty to do a 180 degree turn with 300 - 500 feet to spare if you follow your procedures and get to best glide speed first.  Remember, into a headwind, this IAS will be slightly faster than Vbg.
Now for that glideslope.  The solution is exactly the same as with landing a high performance glider: increase your angle of descent by reducing your lift/drag ratio.  You would go to flaps 30 and/or full forward slip.
A full forward slip in a 172, into the wind, greatly increases the glideslope.  To land, round out, lower AOA by reducing pitch, and hope your brakes work (They should at least once at 30 knots groundspeed even with the grade).  Amazingly, this is one case where the older Cessna 170 taildragger would actually do a bit better by doing a "wheels" landing (without risking a nose wheel strike).
Another option would be to use the other end of the drag bucket by slowing down and increasing AOA, but under windy conditions this is much more risky.
Thankfully, even the steepest road grades (4-8%) work out to less than 5 degrees of slope.
Although the expert pilot may prefer to "stick" the landing uphill, the chances of completely missing a downwind approach make it unadvisable with that much tailwind.

I'd say when the slope of the road is 17% downgrade and the glide path of the plane is only 12,5% the glide path will not meet the slope, and this doesn't even consider the ridge lift effect...

@Timothy Truckle 12.5% is optimal.  Don't forget even before adding flaps the glide angle is much steeper because ground speed is reduced while rate of descent is little changed.  Your point about upslope winds is good, but for the emergency landing in a 172 the lower groundspeed could be favored.  Nice video.

I'm regularly practicing up- and downhill landings at the Wasserkuppe with both, gliders and powered planes. I for myself would never land downhill against the wind on an unknown ground.

well, if it were a road, at 30 knots I might just stay on it even if I couldn't stop ;=).

steepest road grades (4-8%) I'm not sure where you are, but I've cycled plenty of roads that climb more than that, for longer than a typical GA runway - they're just not straight

If there was no wind, I'd probably land uphill, but in the circumstance you describe I'll take the into-wind direction landing down the hill.  This gives the lowest ground speed and therefore the lowest energy state if things don't go so well (you have a 30 knot energy difference between the two landing directions - that's a lot, and a bigger deal all around than the uphill/downhill aspect).
The into-wind glide slope being steeper than the down wind glide slope, the headwind negates some of the problem of having to achieve a steeper decent angle to touch down on a down sloping hill, and you can make a 172 come down pretty steeply anyway if you really have to. If I misjudge the whole thing trying to land into wind down the hill, and don't touch down until the far end, I roll into the bushes doing 40 kts at ground contact, instead of 70 kts if I overshoot going the other way.

I completely agree... In that situation, the main issue is to minimize the ground speed, and that means taking the into-wind direction.


text:
What is a general rule for crosswind correction, especially when landing?
I am wondering what is the rule of thumb with correction of the bat?
Is it 2° for every 5kts?
Now i know you can use a E6B.
But im talking about landing ILS or visuals without bracketing the approach.
So lets say you landing on runway 06 the magnetic heading is 060, the wind is 080/05kts.
So would the correction be 062?
The aircraft is a C172
Apologies if this is not a valid question, getting back into aviation

use the angle that makes you move parallel to the runway...

@ratchetfreak, that is fbf? fly by feels :)

@CGCampbell No, it is trial and error.  The wind is constantly changing anyway, so you need to make constant corrections to compensate for the wind.

the calculation of crosswind correction angles is useful for dead reckoning in cruise. For landing you would fly whatever keeps you aligned with the runway / ILS. The wind you encounter during the approach will typically be different from the reported surface wind.

I thought as much.
Remember chatting to some one and they mentioned the 2° to 5kts.
Thank every one

also, remember that runway 06 could have a heading of anywhere from 055-064, or even beyond - at large airports with multiple runways they sometimes have parallel runways with different numbers. Check the AFD for the actual runway heading :)

@delta Lima, best answer.  You ought to move this out of comments, it would get my vote...

What I have learned during my commercial flight training is making use of the speed number.
Take your TAS, divide it by 60. This is your speed number.
Now take you XWC (crosswind component).
Divide the XWC by your speed number. This is the amount of degrees you should crab to stay on track (wind correction angle)
Lets use an example:
We are flying in a C172 at 120kts TAS.
XWC is 18kts from the left.
120 divided by 60 is 2, so our speed number is 2.
18kts wind divided by 2 is 9. Now adjust your heading by 9 degrees to the left (into the wind), and you should stay on track.
Worked perfectly fine for me so far.
Hope it helps! Cheers

and my answer came to that: $$60*\frac{XWC}{airspeed}=\frac{XWC}{airspeed/60}$$

It helps explain that "divide by 60" if you call "speed number" as "number of miles per minute". ie 120 kts is about 2 miles a minute (ie. 120/60 => 2). makes it easier to remember as "cross wind divided by number of miles a minute"

@Radu094: This may help remember the rule, but it doesn't actually explain anything. (Mathematically the number shouldn't be 60, but 57.3 which is the number of degrees in a radian -- but that much precision is not warranted here).

Well, these days, when every pilot has at least a hand-held GPS unit at a minimum, wouldn't you just steer whatever heading gives you the desired ground track?

@quietflyer Even better, since it gives you your current Track, the difference between it and your heading is the WCA, which you can directly apply to your desired course.

If we set our runway to be aligned to a $x$ axis so the angle is $0°$ and we have an airspeed of $\vec{v_a}$ and a wind of $\vec{v_w}$, this means that ground speed is $\vec{v_g} = \vec{v_a} - \vec{v_w}$.
We want the $y$ component of $\vec{v_g}$ to be 0 so this means that the $y$ components of $\vec{v_g}$ and $\vec{v_w}$ must cancel each other out.
The $y$ component of the wind is our crosswind ($v_c$). To get the $y$ component of our airspeed we take $|\vec{v_a}|\cdot\tan \theta$ where $\theta$ is our heading (0 is parallel to the runway).
This means that $|\vec{v_a}|\tan \theta - v_c=0$ or that $\tan \theta = v_c/|\vec{v_a}|$.
At low crosswind speeds this means that your crab angle in degrees is $\sim 60*\frac{crosswind}{airspeed}$.

Bad at math... if crosswind is 5kts and airspeed is 80kts, 5/80 is .0625, heading is 060, so 60 * .0625 is 3.75, so my crab angle should be ~ 64 degrees?

Woaw! MathJaX equations

@CGCampbell, the 3.75 is your crab angle in degrees. If the crosswind is from the north, then your heading would be 064.

Bleh. Trig. I liked your shorter answer better!

My flight instructor once told me about an incident where an examiner asked him what the correct angle would be for a given crosswind scenario. He simply stated the answer without hesitation. Little did the examiner know that he's a math professor. - lol

I don't see how your last equation falls out. Given $\tan \theta = v_c/v_a$, $\theta = \arctan {v_c/v_a}.$ By Taylor approximation, for small $\theta$ $\arctan x = x$.

I just learned a radian is approximately 60 degrees!

Why would "the $y$ component" of the airspeed vector be $|\vec{v_a}|\cdot\tan \theta$ ? The component of $\vec{v_a}$ perpendicular to the runway heading would be $|\vec{v_a}|\cdot\sin \theta$.

These days, when every pilot has at least a hand-held GPS unit at a minimum, wouldn't you just steer whatever heading gives you the desired ground track?

Turn into the wind so you continue on a course parallel to (better directly over) landing runway. Eyeball. No computation required. As you flare, kick rudder to swing nose over runway centerline as you descend. If you time it correctly, you'll land rolling forward down the runway rather than taxiing off across it.

This is the method preferred by pilots of airliners, while pilots of light planes often prefer to use the wing-down method to have a stabilized no-drift situation all the way to touchdown. Especially pilots of taildraggers.

People are getting way too technical here. You’ll do this kind of the computations for wind correction in cruise flights, but on final approach, that’s all done by visual reference and visually tracking the aircraft down the centerline of the runway during the approach. Once you enter the round out it’s only enough rubber pressure to keep the nose pointed at the far end of the runway, and enough aileron pressure to hold the aircraft on the center line on the runway. That’s it.

Just look out of the window, crab down the centre line and kick off the drift at the flare. No need for numbers.
I gather this doesn’t work so well in airliners.

Welcome to aviation.SE. I fail to see what your answer adds to already existing answers. Could you please expand it so that it is not a simple repetition of already existing answers.

@Graham no it is the opposite, this is the method preferred by pilots of airliners, while pilots of light planes often prefer to use the wing-down method to have a stabilized no-drift situation all the way to touchdown.  Especially pilots of taildraggers.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How is a missile lock detected?
How can a fighter aircraft involved in a face-off detect a missile lock from a trailing aircraft so that it can make evasive maneuvers?

You might find the following article instructive: http://gizmodo.com/how-fighter-jets-lock-on-and-how-the-targets-know-1644871272

I feel like there is a dupe somewhere on this site but i can't seem to find it...

Frequently, one detects a missile lock by getting shot down.

To add some data to Matthew's answer:
Anti aircraft missiles come in basically 4 types (some others have been tried but aren't in common use).

Active radar homing
Passive radar homing
Infrared homing
Laser guided

Active radar homing has a radar in the missile sending out signals. Those signals can be detected and classified by the target aircraft.
Passive radar homing has a receiver in the missile reacting to specific signals bounced off of the target by the launching system. These too can be detected and classified by the target aircraft.
Infrared missiles are completely passive and can't be detected this way.
Laser guided missiles are like passive radar homing missiles, except they react to reflections of a laser beam rather than a radar signal. These too can of course be detected and classified with appropriate sensors.

There have been some attempts at detecting missiles by their own infrared signature, typically their engine exhaust. But this is problematic because most of that is of course blocked from the missile's target by the missile body, and also many missiles will spend a lot of their flight towards the target in an unpowered glide state, thus not having a hot engine exhaust.
Other systems generally can't be detected by the target either. Think optical guidance using a radio link with the missile (while you might be able to detect the link, you can't readily know what it's doing or that you're the target if you do recognise it as a missile guidance link).
Optical guidance using control wires is not generally used with anti-aircraft missiles, but sometimes anti-tank missiles using these systems are used against slow flying aircraft. These are utterly impossible to detect.

I don't believe anti-aircraft missiles -- which travel at supersonic speeds -- spend much of their flight path "in an unpowered glide state"...

@DevSolar you'd be surprised. Most have only seconds of fuel, extremely rapid accelleration to very high Mach speeds, and coast the rest of the way.

I'm surprised. ;-)

@DevSolar so was I when I first read that years ago in a long forgotten book. But it kinda makes sense. You either get rapid accelleration OR long burn time for a given amount of fuel. And rapid Accelleration is prefered unless maybe for very long range weapons where long flight time isn't a problem (and there you'd likely employ a manned interceptor armed with its own missiles or rockets).

So... all the movies showing AAM's trailing contrails (not exactly contrails, but white smoke) all the way are basically Holywood getting it wrong? Again? ;-)

@DevSolar might have been somewhat accurate in the 1950s, but we've long since gone to smokeless engines, current missiles are pretty hard to spot visually, they're designed for it :)

Note that there are also passive radar missiles just using an omnipresent awacs radar signal; in such situations it is not possible to detect if someone locked/fired at you or not.

@PlasmaHH yes, home on radar (or indeed home on any emissions) exists. Same effect as IR homing, can't be detected by the target

You have the terms slightly wrong. Homing on reflections from radar transmitted from the launching aircraft is "semi-active radar", not passive. "Passive radar" is homing on the target aircraft's own radar transmitter.

@RossSmith: I thought those were called anti-radar. Passive radar homes in on radar signals bounced off the plane from background radio emmisions (TV & radio broadcast, cellphone signals, civilian airport radar etc.)

"utterly impossible to detect"... until you explode :-)

@DevSolar Some medium and long-range Air-to-Air missiles only have a few seconds of fuel, and they work by first climbing to a higher altitude than the target and falling back down onto it at high (supersonic) speeds.

The wire-fed anti-tank weapons (Sagar) can be seen coming as they have a very bright trail and they move relatively slow (but much faster than the tank).

@dotancohen yes, but what mainly makes them (next to) useless for use on aircraft are the control wires. The Germans tried is in WW2 with their first AAMs, and it didn't work well.

Infrared missiles are detectable by their actively cooled sensor. If you spot a super cold spot behind and below, this is the infrared missile coming at you. All it needs is to constantly monitor the ground with a high-resolution IR sensor.

@PeterKämpf nice in theory, in practice the spot'd be too small and/or masked by the rocket exhaust heating the missile body. Far easier to detect the exhaust plume, of course that's not going to give you detection all through the flight.

Innovation happens when a new technology is demonstrated to work and all naysayers fall silent. Maybe you don't know what is already possible. The exhaust is hot only for 3-5 s, the sensor can be seen for all 20-30 s flight time of the missile.

May be a stupid question but, taking into consideration that such a missile may be traveling at a speed of X mach. The missile would then have a very high temperature due to air friction. Wouldn't it then be possible to detect the missile using thermal sensing?

The engine exhaust from a rocket motor is highly reflective to radar and may provide and indication to the aircraft of a missile launch.

A general search radar, because it has to search a much larger portion of the area around the plane, can only scan so many times a second.
When that radar finds a target, and the pilot commands the system to lock onto the target, it enables a different radar system, that searches a much smaller portion of the area around the plane where the target is known to be.  This gives not only greater resolution on the target, but it can scan it much faster because it's only scanning a small portion of the area around the plane.
Targeted missiles also only scan a small portion ahead of them, and do so very quickly so they can react quickly to changes in target vector and position.
Most "missile lock indicators" simply listen to how frequently a radar scan takes place, and when it starts happening very quickly it indicates that the faster, more focused radar has found them and is considered locked on, or that a missile with a fast, focused radar has found them and is locked on.

The subject of a radar lock-on may become aware of the fact that it is
  being actively targeted by virtue of the electro-magnetic emissions of
  the tracking system, notably the illuminator. This condition will
  present a heightened threat to the target, as it indicates that a
  missile may be about to be fired at it.

http://en.wikipedia.org/wiki/Missile_lock-on#Detection_by_the_target

Before we talk about lock-on, let's consider the WWII battle between the RAF's u-boat hunting radars and the Metox radar detector on the German submarines.
The RAF introduced their first sub-hunting radar in 1940. This initially consisted of two antennas that had a wide broadcast pattern, about 30 degrees on either side of the antenna centerline. They put one under each wing, pointed outward at 22.5 degrees. Note that this results in an overlap area in front of the nose where both antennas cover. A motorized switch sent the radar signal alternately to each antenna, I can't recall the exact speed, but for argument's sake, let's say 100 times a second.
When the aircraft was searching for targets, in most cases it would only be visible to one antenna or the other. At first the aircraft would fly past it until they saw the signal fading, meaning it was passing about 60 degrees off the nose. The navigator would then plot a likely position, and the aircraft would turn towards the plot.
At this point, the uboat would be visible to both antennas. By comparing the strength of the signal in the two, they could tell which way to turn until they were pointed right at it.
The Germans figured out what was going on when their uboat losses shot up in early 1942. They responded with the Metox detector, simply a radio receiver tuned to the RAF's radar frequency around 176 MHz.
Now imagine what this is like for the radio operator listening to Metox. When the aircraft was still searching they would hear only the signals from one of the two antennas, so it would warble in their earphones at the switching rate - in this case they would hear it buzzing at 50 Hz. When then aircraft turned towards them they would start to hear the signal from both antennas, so the tone would suddenly jump to 100 Hz. They knew the aircraft was now approaching them, and would dive.
By this point the British had figured out that they could compare the signals strengths from the two antennas electronically, much more accurately than a human could. This coincided with new radars in the microwave region that needed antennas only a few cm long. Now it became very easy to put two antennas right beside each other and let the electronics figure out which was closer to the target. The output was the "error signal", which was amplified and sent to motors pointing the antennas, and this caused the entire platform to automatically track the target. Once again, the target could tell it was being tracked by listening to the signal. If it was pulsing and then suddenly became steady, the radar was locked on.
Additionally, some radars changed the pattern of the entire broadcast. This is often used to put more signal on the target during a dogfight or while a missile is being fired. For long-range work radars tend to put out a smaller number of longer signals, while at shorter ranges more signals of shorter duration are better. The radar detector can listen for these changes to indicate lock on. This was the method used by the USAF's systems over Vietnam, they listened for a change in the pulse repetition frequency of the SA-2's radar.
Modern radars and missiles don't do this, and detecting lock-on is basically impossible now. Very modern AESA radars generate different frequencies and signals with every pulse, so a receiver doesn't get the same signal twice. This makes it almost impossible to know a radar is even painting you, let alone that its tracking you. Additionally, the missiles don't track continually, instead they receive an initial location from the aircraft's computer and then fly to that point in space, then turn on their own radar. The target generally doesn't know a thing until the missile goes active a few seconds from impact.
This is why the UV detection is so important, even though, as others have noted, it is not terribly effective.

The missile motor has a UV signature that can be detected.  Newer IR sensors can be fused to a UV sensor and generate a warning.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why do jet engines sometimes spit out flames after ingesting birds?
After seeing this video of a 757's engine catching fire because of a bird,
I wonder:

What happens to the engine?
Why does it periodically exhaust fire?
Is this some safety feature?

I guess that is a compressor surge cycle initiated by the ingestion of the bird and associated damage. The fire is presumably from an excess of unburned fuel due to the pilots/FADEC not cutting engine power at take-off. It looks more like a mode of natural oscillation rather than a safety feature!

Note that the engine is not really on fire. The flame is only in the normal flow path and will stop once the fuel is cut off.

Just as a matter of note, you can also get a compressor surge if you use excessive reverse thrust for the forward speed that you have, and especially as the engine nears time for overhaul.

Because birds are flammable? ;)

As RedGrittyBrick has commented, what happened in that video is a compressor surge. This is not really an engine fire, but a temporary backfire. To understand what is happening, look at this diagram of a turbine engine:

The air is compressed in the front of the engine, fuel is injected and ignited, and then everything flows out the turbine section and the exhaust nozzle. The only thing supporting the compression at the front of the engine is the rotation of the compressor blades. There are many different things that could cause the compression to fail. In the event of a bird strike, these blades can be damaged.
When that happens, there is highly compressed air in the combustion section, and there is no longer enough compression to prevent it from exiting the front of the engine. This causes the air to exit both the front and rear of the engine. The disruption in air flow happens too quickly for the engine control systems to react. This means the engine is still pumping enough fuel to run the engine at normal thrust, but there is no longer enough air flow to burn all that fuel in the combustion section. The result is that the fuel burns through the turbine and exhaust section of the engine, and behind the engine.
Once this compressed air has exited the engine, it will generally recover. The engine is not on fire, so once the excess fuel burns away, the flames are no longer visible. However, if the problem is still present (and damage from a bird strike would not go away), then the engine surges over and over until it is shut down or set to a lower, more stable power setting.
So this is not so much a safety feature as physics taking over, and there is no real fire of immediate danger to the airplane. The engine system will automatically shut down the engine if it senses there is anything potentially very dangerous or that completely prevents the engine from running. While compressor surges can eventually damage the engine, it's not really a catastrophic event, and it is up to the pilots to assess the situation and shut the engine down if necessary. Reducing power to idle may sometimes be sufficient to stop the surges.
During a compressor surge, the flames are confined within the engine. There should be nothing to burn here besides fuel, so cutting off fuel will stop the fire. Actual engine fires are generally caused by flammable fluids leaking into hot areas around the engine, or the hot engine gases escaping from the air flow sections into surrounding systems. There are detection systems to alert the pilots of this situation, and the engine area is designed to contain a fire until the pilots can use the fire extinguishing systems to put such a fire out or at least get the plane back on the ground.

I would stress that the engines is not really on fire during compressor surge/stall. When fuel is cut off, the engine just flames off and stops and no extinguishing is needed.

Also, the engine in the video did show an oscillating surge, alternately compressing air almost like in normal operation, and then much less. Since the turbine is sort of blocking the flow path out back, it needs a minimum compression performance to work, and with the damage this performance was no longer possible. Once the compressor pumped air back, this air would not exit quickly enough through the turbine, but would partially flow back through the compressor. Now the power to keep the compressor turning decreased, the flow went back to normal, and the cycle repeated.

So how would a pilot not looking at the engine to see the fire detect surge from his instruments in order to take corrective action e.g. reducing power to idle etc. Is there a surge warning alarm or something like it in the avionics?

@curious_cat That sounds like a good separate question.

When turbine engines ingest birds or other objects the result is Foreign Object Damage (FOD).  The metal parts of these engines are operating under extreme stress very close to the absolute limits of what metallurgical technology can produce.  The flames you see are bits and pieces of the object coming out of the engine along with bits and pieces of the engine.  Think Capt. Sullenberger and miracle on the Hudson or Capt. Al Haynes and crew with UAL 232.  Turbines are designed to have only two things pass through them air (of various temperatures) and fuel.  Anything else causes problems up to and including total failure.
Hope this helps!

UAL 232 was a fan disk disintegration due to metal fatigue, resulting in loss of hydraulics and subsequent loss of controllability. There are very few birds, and not really a whole lot else either, at FL370.

My point is that anything (other than air or fuel) passing through a turbine will mostly likely cause damage and flaming debris to exit the engine.  Also, it may cause the engine to fail in a complete self destructive manner.  Birds have been encountered at some incredible altitudes (not FL370) but ash and pumice from volcanoes have damaged windows and engines at similar levels.  Jon

@JonCaples: Actually, birds have been encountered at FL370.  They were found by - guess what - getting sucked into a jet engine.


text:
What is the nominal frequency of the satellite communications terminal on MH370?
What is the nominal frequency of the aircraft terminal on flt 370? (The terminal used to provide the handshake with the satellite). The raw data provided by the accident team does not specify this frequency.  This frequency info would help in checking their BFO calculations.

I looked through the report about the flight path analysis that is available from this page.
In Appendix G they explain in more detail how the analysis was done. On page 56, they provide the details about the frequencies (emphasis added):

The uplink and downlink Doppler may be calculated from the relative movement of the aircraft, satellite and GES using the signal frequencies of 1646.6525 MHz (uplink) and 3615.1525 MHz (downlink).

This means the aircraft terminal transmits (uplink) on 1646.6525 MHz and receives (downlink) on 3615.1525 MHz. However, it is not that simple.

The aircraft terminal adjusts its’ transmit frequency to compensate for the Doppler induced on the uplink signals by the aircraft velocity. Aircraft heading and ground speed are used to calculate the Doppler shift the signal would experience if the satellite was at its nominal location over the equator.

This is called Automatic Frequency Control (AFC) and Appendix G goes into more detail about all the factors involved, and provides tables with example calculations.

so, is the aircraft terminal frequency 3615.1525Mhz?

@JohnMauch for receiving data from the satellite yes.  only the transmit portion would be relevant to geolocation from the satellite's point of view, however.


text:
What do the terms "Landing Since New", FC, and FH mean on a Landing Gear RDAS?

Could anyone able to tell me the "Landing Since New" Meant? and what is the FC and FH in here?
Thanks

Why would 29,253 Landing Flight Hours be "not relevant"?

Landing Since New - The definition is probably provided in Page 2 of the RAS.
This basically means the number of hours spent Landing from the time the aircraft was brand new . Based on this figure the blades will have to be removed.
FH - Flight Hours - Basic unit for measuring length of time in service/before maintenance becomes due. The flight hour (FH) represents one hour of flight
FC - Flight Cycles - One Take Off and Landing

The number of hours spend landing?  How would you even determine this?  That is not what Landings Since New means.

Landing Since New
This is the number of times that this nose gear has been used while installed on an aircraft when it landed.  I.e. the number of landings that it has been exposed to since the date that it was manufactured (new).
FC
The number of cycles (takeoffs/landings) that the aircraft from which the nose gear was removed from had at the time that it was removed.
FH
The number of hours of flight that the aircraft from which the nose gear was removed from had at the time that it was removed.
Note that the aircraft FC and FH don't really mean much in relation to the nose gear unless it has been installed on that aircraft since the time that they were both new.


text:
Is a 40-year-old person too old to start an ATC career?
My wife:

is 40 years old
is a mother of two (able to do five things simultaneously / in the same time)
has been working as an English and French teacher for her whole career (19+ last years)
has very little to do with airline industry (being a passenger and low-level enthusiast)

I think that this is enough (especially point 2. and 3.) and I'd like her to at least try to become an ATC (i.e. enroll for a preliminary English test).
She is reluctant saying that she:

is too old
have very little or no technical background / education / experience

to pass even first tests, not mentioning becoming a professional ATC.
Can anyone judge which one of us is right or at least provide some source of information, based on which such judgement can be done? I.e. to formulate an answer (similar to this one) on unique challenges and limitations for starting a career as an ATC at 40 years old?
It is possible in the country where we live (Poland). The original limitation (26 years old) was raised to 35 years old some 10-15 years ago and then 40 years at most. And now the age barrier is lifted completely.

Is this something that she wants, or that you want for her?  (I know, more of a marriage question...)

Well she has to want it pretty bad to even get anywhere in that trade, so it'll get resolved pretty quickly.  I'm glad I didn't make it past the first stage in the end.  Being a highly creative person, the "repetitive" nature of the job (not sure how else to describe it) would've bored me to tears once past the learning curve.  The biggest problem is if you dump a job to start through the training, the risk of washing out before finishing is very high (I was told only about 20% of initial applicants accepted made it through), leaving you on the street.

I'm guessing item 2 on the first list is at least somewhat humorous, so I won't bother digging up scientific studies about multi- vs. switchtasking.

Something to check on is the medical requirements.  I don't know about Poland, but in the US, a controller needs (I believe) a class 2 medical, the same as a commercial pilot.  The older we are, the more medical issues tend to crop up.

Some years ago (in the 80s) I tried out for ATC in Canada and took the preliminary screening test, which was a series of 25 diagrams representing radar displays with targets moving around different airways, crisscrossing each other in various directions, and you had to answer 2 questions for each diagram, whether or not targets at different speeds and altitudes would conflict with each other (there was a table with all the targets' speeds, tracks and altitudes).  It was a timed test and I wasn't paying attention so I was at question 45 when time was called and you had to drop your pencil right away.  So I didn't answer the last 5.
I got one or two wrong, giving me a score of 43, a "pass" for the test, but there so many applicants they were only taking candidates with scores over 45 for the next stage.  Probably would have made it if I'd just answered the remaining 5 at random.
Anyway, actual background is not that important. They are looking for a very special set of mental skills, and someone has them or they don't.  Mainly, a very high cognitive and spatial IQ, and a very powerful memory.  To the extent that educational background matters, it would mostly be as a sign of intelligence.
I used to know a centre controller who told me that once trained and experienced, and having the required mental abilities in the first place, and the fact that the high stress periods don't last an entire shift (there are "rush hours" where things are intense, then relatively quiet) the job wasn't nearly as tough as the public perception.
She should just go ahead and apply and start the screening process, as I don't believe there are any specific professional or educational prerequisites, and there is no harm in trying.  She will probably find out pretty quickly whether there is a chance.  One thing going for her is the overall personnel supply situation which is affecting ATC to some degree as much as other parts of aviation (shortages of applicants with what it takes).

Thank you for a detailed and comprehensive answer.

There is a huge difference between today’s teaching methods and those of 80’s. It isn’t necessary that you will always have rush hours and stuff as it totally depends at which airport you work and how highly reputed(for being busy)it is.

Nothing to do with teaching methods.  I'm talking about filtering candidates inborn mental skills. Of course there are control towers where the controller sits around doing nothing 90% of the time.  The rush hour stuff happens at area centers and ATC units at major airports.

It is country specific
In the US, as well as I can translate the legalese to English, your application must be accepted by the FAA before you turn 30. (I considered it myself, about a decade too late.)
In Canada, you must be at least 18 years of age, but I couldn't find any readily available maximum. As another answer indicates, there may not be one.
In Denmark (per a comment by J. Hougaard), there used to be a maximum age for applicants, but it was removed a few years ago because it is against EU Law to discriminate based on age. He assumes the same applies in other EU countries, including Poland.
That said, in Poland, PANSA appears to be the agency involved. I can't translate any of it, so you're on your own there.

In Denmark there used to be a max age for applicants, but it was removed a few years ago because it is against EU Law to discriminate based on age. So I assume the same applies in other EU countries, including Poland.

I was told that age limitation was lifted off in Poland due to very few (too few) people applying for each bi-yearly application period. Your comment (about discrimination) is a very good shot to consider on the other hand. But, then it would not explain why current period's end time for application has already been extended two times. Thus, I assume that a very limited number of people applying here in Poland can again be at stake.

For the USA it is 30 or under, that is before 31.

I think that a lit bit technical knowledge is required  to at least think of being an ATC because imagine a “mayday” situation. The ATC has to plan everything according to it and this is the point where technical knowledge comes really handy.
 Secondly it isn’t necessary that doing five tasks together at the same time makes you suitable for an ATC because there can be no room for error and being one is not as easy as one may think.
 Also,Age won’t be a limiting factor but maybe you should think about your kids as well,being an ATC would require lots of training hours and after that maybe 6-8 hours of shifts per day that would stress her out mentally and physically which may have some effects on the upbringing of your kids. If you think that her job is necessary for your financial requirements then you could try but let it not affect your kids(that is the worst thing specially if they are about to be teenagers).

in the Netherlands, LVNL will not hire anyone over the age of 30 (this used to be 25 until some years ago).
It's listed as the very first sentence on their website about how to get a job as a controller.
There are also strict medical requirements basically the same as those for a PPL class medical.
This is in no small part because there's a 5+ year paid for education prior to becoming a controller and the agency of course wants to get their investment back. And if you're much older you're likely to not be able to work the high stress job long enough for that to happen, it's that simple (as was explained to me by a recruiter when I asked about the age restrictions years ago).

Do you have a source to confirm the fact that age limitation still is in force in Netherlands? It would contradict comment given to the other answer (about age limitation lifted off across the whole UE due to discrimination). Netherlands is still part of the UE as far as I remember! :>

@trejder I just checked their website and the 30 year age limitation is still there.


text:
How does a variable-sweep-wing aircraft keep wing-mounted payloads straight?
So, a variable-sweep wing is an interesting design, but I am wondering: if the aircraft wing moves back or forward, then the payload's angle will also change. Is there a another rotating mechanism in the payload rack or wing to keep the payload straight?
For example, as demonstrated in this picture of a Tornado ADV  where the red lines are payloads. When the wing goes back it also changes the angle of the payload, unless there is another mechanism to keep the payload straight.
However, some might argue that payload isn't carried on the variable part of the wing, as on the F14 (See the link for reference).
But, there are numerous examples of aircraft that do carry the payload on the variable section of the wings, such as the Panavia Tornado ADV and Su-24 (see the links for references).

I was able to find this Panavia Tornado cutaway picture (took about 5 seconds):

(Air International 1984-09 / P.Jackson - Tornado)
Inside the left hand wing you can see a part labelled 197, with notation "Swivelling pylon control rod".
This rod is tied to the main body of the plane, and while the wing changes its position, the rod pushes the pivoting pylon to maintain pylons straight alignment.

Yes, that is the simple pantograph mechanism I referred to in my answer.

Mechanically this can be done using the simplest form of pantograph, as a hinged parallelogram with one end fixed to the fuselage and the other to the ordnance pylon. Or, I suppose you could implement a digital one.

Thanks for your answer, however, are you making a logical assumption or do you have any evidence to support your answer, such as a image showing the "Pantograph" on the wings?

@BilalMirza Not "on" the wings but "in" the wings, these planes are highly streamlined! Someone else has posted an actual example.

So, the probable exploitation was that the payload is rotated to keep it straight. However, at first, I haven't found any photo/info on google to support that explanation.
But, I have found the answer to my own question. You can see the scratch marks generated by the movements of the racks (possibly pantograph) on the Panavia Tornado aircraft (for full size picture see the link1 and link2).
Those marks are not visible on majority of pictures available on the internet, I have found these pictures after spending a lot of time.

I have provided you with proof of the "pantograph" in my own answer.


text:
Low-wing taildraggers pictured on 2021 ASA FAR/AIM cover?
There are a pair of low-wing taildraggers on the cover of the 2021 FAR/AIM published by ASA. Can anyone identify them?
I tracked down the image source here, but the description does not identify the planes:
https://www.istockphoto.com/photo/motor-aircraft-on-the-ground-gm1131409963-299576828

Zwei farbige Motorflugzeuge Sportflugzeuge stehen am Boden auf dem Rollfeld eines Flugplatzes.
Translated to English:
Two colored motorized aircraft, sport aircraft, are standing on the ground on the tarmac of an airfield
Does anyone know what they are?

Those airplanes are Jodel1050 Ambassadeur. The red one is D-EHKA. The picture has been taken at EDHS, a former military airfield built by the Luftwaffe in WW2.
D-EHKA is based in EDXR (northern Germany)
It has been built in 1964 in France and still flies 100-150 hours every year.
Very efficient, o200A, 5 Gal/h and 100 kts cruise. Can fly up to 6 hours.
Best regards,
Christian
(no idea how this picture made it on the ASA FAR AIM :-)

They are Jodels. All wood.  The D-11 was a very popular homebuilt once upon a time (a friend owns one), along with factory production of various models in France in the '60s.  Hard to say exactly which models those are.

Thanks, @John K!

Are you sure? They look a LOT like Robin DR.221s...


text:
My N number showed up on another plane's flight on FlightRadar24. How does that happen?
My Cessna 150F has never had ADS-B out. It has had the same N number since 1967. But yet, FlightRadar24 shows a flight a few weeks ago, perfectly tracing the coast north of San Diego, 2,000 miles from my home airport in Kentucky.

How is it possible for ADS-B tracks to show up under the wrong N number?
It is not one blip of data with a bad ID…it’s hundreds of data points over 50 minutes with no discernible glitches at all.
I searched the web and this StackExchange for several variations of “ads-b wrong plane” with no results coming close to matching. Plenty of discussion of poor position accuracy and other anomalies…but none about wrong N numbers that I found.
UPDATE: I found the name for this, which yields better search results: "Call sign mismatch" or its abbreviation, CSMM.

At a guess I'd say that the ADS-B unit in the aircraft you saw hasn't been configured properly (perhaps it's a new installation). It shouldn't happen, but people are fallible. There's a good article on I Fly America that discusses it.

I assume you didn't leave your keys in the ignition?

FR24 may not actually receive any N-number from the aircraft. It's not unlikely that it only receives the Mode-S transponder code and cross-references that with a database to your N-number.
So the real question is, how is it possible that another plane broadcasts the same Mode-S code? There's really only two options here:

a receiver was set up incorrectly. FR24 is based on a network of volunteers, so perhaps one had a faulty setup. Or there was a bug at FR24.
the transponder was configured incorrectly. Somebody has to put in the code in the transponder. Perhaps they made a mistake. Perhaps the transponder had a fault. I think this is the more likely option.

Seeing as no further flights are logged on FR24 since then with your N-number, either way the problem seems to be rectified.

I would not exclude the possibility of a bug in any of the software involved in the FR24 network.

An incorrect receiver setup is unlikely to be the cause. I can't think of any probable setup mistake that could cause this, since the 24 bit address is so fundamental to the way ADS-B works. I would rule it out. The transponder being configured incorrectly, or having a hardware issue (in case of hard-wired 24 bits), is a common cause of 24 bit confusion.

@DeltaLima Agreed, but as bogl said, bugs happen.

@DeltaLima Since the OP's aircraft has never had ADSB-out, how could any misconfiguration cause his N# to come up? His N# being in the database matched up against any 24-bit code seems like an error in & of itself, yes?

@RalphJ Mode-S is not the same as ADS-B

"FR24 does not actually receive any N-number from the aircraft", why not? ICAO states about the Flight Id field: "The aircraft identification (sometimes called the flight identification or FLTID) is the equivalent of the aircraft callsign" and AOPA says: "for most general aviation operations, the flight ID is simply the aircraft’s N number." This field is entered manually, hence errors are possible.

I used to develop air traffic software.  I remember seeing a case where 2 aircraft at the same airport were broadcasting the same Mode-S at the same time.  The explanation I was given was it was probably degradation of the memory in the transponder, likely a single bit flipped from 0 to 1 or vice versa.

@RalphJ The FAA has an alhorithm that assigns 24 bit addresses based on the registration. If you know the algorithm you can relate any US registration to its 24 bit address and vice versa. I am pretty sure that FR24 uses this algorithm to convert US 24 bit addresses that are not in their database to the registration nr.

Reminds me of the video where the protagonist flew into a Class C airspace, *hearing" all the appropriate radio authorizations.  Turns out, spoiler alert, the clearance was being issued for a very similar call sign, and this aircraft had a broken radio antenna.

Flight tracking sites like FR24 use both ADSB data and official radar tracks.
The first possibility is that someone misconfigured their ADSB to use your hex code/tail number. This seems unlikely since setup is normally done by an avionics technician, but it’s not impossible. It becomes more likely if the plane is used for operations like Angel Flight, where the pilot has to frequently change their own ADSB setup due to using mission callsigns, which is unfortunately not a user-friendly operation with most systems.
The other possibility is that the pilot called ATC for flight following, and the controller heard or typed the wrong tail number when creating their beacon code. Controllers are human too and do make mistakes occasionally.

I've noticed this often locally where I'll see an ADS-B track with the correct tail number for an airplane I recognize and a duplicate one (presumably from official radar tracks) that has a smaller "typo" version of the registration.

ADS-B codes in the US are calculated based on the N-number.  If you ever get a Mode S transponder for your airplane, your ADS-B transponder code will be "AC534B".  That same calculation can be run in reverse.  If someone sets the code "AC534B" in their transponder, they'll show up on FR24 and other flight-tracking sites as "N8933S".
ADS-B transmissions can also include an optional "callsign" field up to eight characters long.  Flight-tracking sites tend to display this in preference to the tail number because knowing that SWA 1691 is on final approach to Los Angeles is more useful than knowing that N7879A is on final approach.  If someone set their callsign to "N8933S", it will be displayed instead of the actual tail number.

Radarbox is willing to say that the track did belong to a transponder with the code AC534B, not a callsign, and that they received the data from FAA SWIM, not a private ADS-B receiver.

It seems many ADS-B transponders lets the pilot reconfigure their callsign in the unit: https://www.aopa.org/news-and-media/all-news/2017/october/pilot/adsb-changing-flight-id
This is news to me. I was not aware this is pilot-configurable.
This is typically done for a special kind of flight, such as Medivac, Angel Flights, or scheduled commercial flights. This allows the aircraft ID in the ADS-B data stream to match the ID filed on a flight plan. In fact, it seems that it is required that they match.